这里只是记录一下,非常推荐马同学高等数学,文末有原文.点击这里看里面的例一应该是理解贝叶斯公式最好的例子
,如果你稍微有一些基础,我觉得文末第二个链接中的例一更加适合你

代数推导

1. 贝叶斯公式

是根据条件概率推导的
P(A|B)=P(AB)P(B)P(B|A)=P(AB)P(A) P ( A | B ) = P ( A B ) P ( B ) P ( B | A ) = P ( A B ) P ( A ) <script type="math/tex" id="MathJax-Element-1"> P(A|B) = \frac{P(AB)}{P(B)} \qquad P(B|A) = \frac{P(AB)}{P(A)} </script>
所以推导可以得到Bayes公式:

P(A|B)=P(B|A)P(A)P(B) P ( A | B ) = P ( B | A ) P ( A ) P ( B )
<script type="math/tex; mode=display" id="MathJax-Element-2"> P(A|B) = \frac{P(B|A)P(A)}{P(B)} </script>
通常会把P(B)看成归一化系数 η η <script type="math/tex" id="MathJax-Element-3">\eta</script>;
P(A|B)=P(B|A)P(A)P(B)=ηP(B|A)P(A)η=P(B)1=1AP(B|A)P(A) P ( A | B ) = P ( B | A ) P ( A ) P ( B ) = η P ( B | A ) P ( A ) η = P ( B ) − 1 = 1 ∑ A P ( B | A ) P ( A )
<script type="math/tex; mode=display" id="MathJax-Element-4"> P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \eta P(B|A)P(A) \\ \eta = P(B)^{-1} = \frac{1}{\sum_{A}P(B|A)P(A)} </script>
这里有什么过程不懂可以看文末的知识补充

A一般是某种状态,B一般是某种观测值
1.P(A) 1. P ( A ) 先 验 概 率 <script type="math/tex" id="MathJax-Element-5"> 1. P(A)先验概率</script>(Prior probability)
2.P(A|B) 2. P ( A | B ) 后 验 概 率 <script type="math/tex" id="MathJax-Element-6"> 2. P(A|B)后验概率</script>(Posterior/causal probability)
3.P(B|A)P(B)Likelyhood 3. P ( B | A ) P ( B ) 可 能 性 函 数 ( L i k e l y h o o d ) <script type="math/tex" id="MathJax-Element-7"> 3. \frac{P(B|A)}{P(B)}可能性函数(Likelyhood) </script>

2. 递归贝叶斯

先简单推导一下三次变量:

P(x|y,z)=P(y|x,z)P(x|z)P(y|z)=P(z|x,y)P(x|y)P(z|y) P ( x | y , z ) = P ( y | x , z ) P ( x | z ) P ( y | z ) = P ( z | x , y ) P ( x | y ) P ( z | y )
<script type="math/tex; mode=display" id="MathJax-Element-8">P(x|y,z) = \frac{P(y|x,z)\, P(x|z)}{P(y|z)} \\ \qquad \quad= \frac{P(z|x,y)\, P(x|y)}{P(z|y)} </script>
这里y,z是可以互换的
如果y,z符合Markov属性,那么还可以推导如下:
P(x|y,z)=P(y|x)P(x|z)P(y|z)=P(y|x)P(x|z)P(y|open)P(open|z)+P(y|open)P(open|z) P ( x | y , z ) = P ( y | x ) P ( x | z ) P ( y | z ) = P ( y | x ) P ( x | z ) P ( y | o p e n ) P ( o p e n | z ) + P ( y | o p e n ¯ ) P ( o p e n ¯ | z )
<script type="math/tex; mode=display" id="MathJax-Element-9"> P(x|y,z) = \frac{P(y|x)\, P(x|z)}{P(y|z)} \\ \qquad \qquad = \frac{P(y|x)\, P(x|z)}{P(y|open)P(open|z)+P(y|\overline{open})P(\overline{open}|z)} </script>
P(x|z1zn) P ( x | z 1 … z n ) <script type="math/tex" id="MathJax-Element-10">P(x|z_{1} \dots z_{n})</script>?
把Z1,…..,Zn看成一个整体,再根据马尔科夫条件,在x已经知道的情况下,Zn同{Z1,…,Zn-1}无关,所以

P(x|z1zn)=P(zn|x,z1,,zn1)P(x|z1,,zn1)P(zn|z1,,zn1)=P(zn|x)P(x|z1,,zn1)P(zn|z1,,zn1) P ( x | z 1 … z n ) = P ( z n | x , z 1 , … , z n − 1 ) P ( x | z 1 , … , z n − 1 ) P ( z n | z 1 , … , z n − 1 ) = P ( z n | x ) P ( x | z 1 , … , z n − 1 ) P ( z n | z 1 , … , z n − 1 )
<script type="math/tex; mode=display" id="MathJax-Element-27"> P(x|z_{1} \dots z_{n}) = \frac{P(z_{n}|x,z_{1},\dots,z_{n-1})\, P(x|z_{1},\dots,z_{n-1})}{P(z_{n}|z_{1},\dots,z_{n-1})} \\ = \frac{P(z_{n}|x)\, P(x|z_{1},\dots,z_{n-1})}{P(z_{n}|z_{1},\dots,z_{n-1})}</script>
所以可以得到递归贝叶斯公式
P(x|z1zn)=P(zn|x)P(x|z1,,zn1)P(zn|z1,,zn1)=ηnP(zn|x)P(x|z1,,zn1)=ηnP(zn|x)ηn1P(zn1|x)P(x|z1,,zn2)=η1ηni=1nP(zi|x)P(x) P ( x | z 1 … z n ) = P ( z n | x ) P ( x | z 1 , … , z n − 1 ) P ( z n | z 1 , … , z n − 1 ) = η n P ( z n | x ) P ( x | z 1 , … , z n − 1 ) = η n P ( z n | x ) η n − 1 P ( z n − 1 | x ) P ( x | z 1 , … , z n − 2 ) = η 1 … η n ∏ i = 1 … n P ( z i | x ) P ( x )
<script type="math/tex; mode=display" id="MathJax-Element-28"> P(x|z_{1} \dots z_{n})= \frac{P(z_{n}|x)\, P(x|z_{1},\dots,z_{n-1})}{P(z_{n}|z_{1},\dots,z_{n-1})} \\= \eta_{n}\,P(z_{n}|x)\, P(x|z_{1},\dots,z_{n-1}) \\ \qquad \qquad \; \; \;= \eta_{n}\,P(z_{n}|x)\,\eta_{n-1}\,P(z_{n-1}|x)P(x|z_{1},\dots,z_{n-2}) \\ = \bf{\eta_{1} \dots \eta_{n}\prod_{i=1\dots n}\,P(z_{i}|x)\,P(x)} </script>

3. 贝叶斯滤波

这里写图片描述

直观理解

这里写图片描述
这里写图片描述

知识补充

这里写图片描述

  • 连续情况下:
    P(x|μ)=P(x|μ,x)P(x)dx P ( x | μ ) = ∫ P ( x | μ , x ′ ) P ( x ′ ) d x ′
    <script type="math/tex; mode=display" id="MathJax-Element-13"> P(x|\mu) = \int\,P(x|\mu,x')P(x')dx' </script>
  • 连续情况下:
    P(x|μ)=P(x|μ,x)P(x) P ( x | μ ) = ∑ P ( x | μ , x ′ ) P ( x ′ )
    <script type="math/tex; mode=display" id="MathJax-Element-14"> P(x|\mu) = \sum P(x|\mu,x')P(x') </script>
参考

https://www.matongxue.com/madocs/301/
https://www.cnblogs.com/ycwang16/p/5995702.html
https://blog.csdn.net/qq_30159351/article/details/53395515

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